Question

1/sinA+cosA+1 + 1/sinA+cosA-1= secA+cosecA

Answer 1

sin a ( 1+ sina / cos a )  + cos a ( 1+ cos a / sin a )

sin a ( (cos a + sin a ) / cos a )  + cos a ((sin a + cos a) / sin a )

(cos a + sina )(sin a / cos  a )  +  (sin a + cos a )(cos a / sin a )

(cos a + sin a)(tan a ) + (cos a + sin a)(cot a )

taking cos a + sin a as common we get

(cos a + sin a )(tan a + cot a )

(cos a + sin a )((sin a / cos a)+(cos a / sin a ))

(cos a + sina )((sin2a + cos 2a / sin a cos a))

(cos a + sin a )( 1 / sin a cos a)

(cos a + sina ) /  sin a cos a

(cos a / sin a cos a)  + (sin a / sin a cos a )

( 1 / sin a ) + (1 / cos a )

cosec a + sec a 

hence proved